Between proper integrals and improper integrals

Question

I just started learning about improper integrals. Many of them are improper because the function evaluates to infinity at some point in their domains, e.g. $f(x)=1/x$ on the domain of $(0,1)$. My question is :

Is there any function integrable on $(0,1)$ with $f\rightarrow \infty$ as $x\rightarrow 0^+$?

The answer might be no, because $f$ is unbounded. But I am not sure.

Answer 1

Consider the function $f(x) := \frac{1}{\sqrt{x}}$. Clearly, $f$ blows up near $0$ but $$ \int_0^1 f(x)\,\mathrm{d}x = \int_0^1 \frac{1}{\sqrt{x}}\,\mathrm{d}x = 2. $$ This function is not Riemann integrable because it is unbounded near $0$. However, it makes sense as an improper and Lebesgue integral. In fact, by the monotone convergence theorem, the improper and Lebesgue integrals agree.

Answer 2

Sure, consider $f(x) = x^{-\frac{1}{2}}$ or, more generally, $f(x)=x^{-\alpha}$ for any $0<\alpha<1$.